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100 years old single malt 17-cell Sudoku

 
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someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Sun Aug 28, 2005 7:55 am    Post subject: 100 years old single malt 17-cell Sudoku Reply with quote

Hi,
This one is a "Leckerbissen". You should enjoy it the same way you drink a 100 year old single malt Whisky.

After 36 numbers (gulps) you get to the top of of the EverEst (or EverWest). The G point.

Here you can apply one of the "advanced" technique. A LOGICAL one.
No need of try and error.
Which one you have to find out by yourself.

I wish you a Good look ;-)

see u,

P.S. I almost forgot the drink for you:

050400000
000030800
000000001
300080700
060000050
000200000
000506040
108000300
000000000

P.P.S. you will need ABSOLUTE no other technique up the the top!!!
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sun Aug 28, 2005 6:30 pm    Post subject: Reply with quote

Well, I'm not sure I used the right technique, but I have solved this puzzle.

After 36 moves I arrived at this point:
Code:

2/7/9   5     1      4   2/7/9   8     2/9    3     6
  6   2/7/9 2/7/9    1     3     5      8   2/7/9   4
  4     8     3     7/9    6   2/7/9    5   2/7/9   1

  3    2/9    5      6     8     4      7     1    2/9
  8     6   2/7/9    3    7/9    1      4     5    2/9
 7/9    1     4      2     5    7/9     6     8     3

2/7/9   3   2/7/9    5    ***    6    1/2/9   4     8
  1    ***    8     7/9  2/4/7 2/7/9    3     6     5
  5   2/4/7   6      8    ***    3    1/2/9  2/9    7

(The squares marked *** all have four possibilities.)

I concentrated on the squares marked with either a {2, 9} or a {7, 9} pair and deduced that one must place a "9" in r4c9 by the following fairly simple logic.

Assume that r4c9=2. Then we have two chains of inference:

r4c9=2 ==> r4c2=9 ==> r6c1=7
r4c9=2 ==> r5c9=9 ==> r5c5=7 ==> r1c1=7

But we can't have two "7"s in column 1, so r4c9 can't be a "2"; it must be a "9."

The rest of the puzzle certainly wasn't trivial, but it went pretty quickly. dcb

PS If anyone else wants to try this one, your first move is "8" in r7c9. I found it was pretty easy to place all the "5"s, "8"s, "6"s, and "3"s, in that order. After that it got a little sticky.
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someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Mon Aug 29, 2005 12:32 am    Post subject: Reply with quote

Hi,
Nice solving! Congratulations!

Still if you take a look at the number 7 and only on it. You will find easy that there are 3 columns that have it twice. And the numbers are in rown in this 3 columns only 2 times (columns 2, 4 and Cool.

This is a Swordfish (on Column) situation, so we can deduce:

7 not in r2c3, it is in r2c2 r8c2 r3c4 r8c4 r2c8 r3c8 (Swordfish on Column)

Did you know about the Swordfish?

see u,
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Guest






PostPosted: Mon Aug 29, 2005 1:19 am    Post subject: Reply with quote

someone_somewhere wrote:
...

This is a Swordfish (on Column) situation, so we can deduce:

7 not in r2c3, it is in r2c2 r8c2 r3c4 r8c4 r2c8 r3c8 (Swordfish on Column)

Did you know about the Swordfish?

see u,


I read a little about it on a couple of web sites, but I haven't yet learned to recognize it easily. I will study this example and maybe learn to see one when it is staring up at me! dcb
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someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Mon Aug 29, 2005 5:25 am    Post subject: Reply with quote

Hi,
I scan the lines(rows) and look for the same number to appear only 2 times.
If I find 3 such lines, than I check that every 2 lines(rows) have the number in one common in only 1 column.

. . . x . . . . x
. .
. x . x . . . . .
. . .
. x . . . . . . x

If I have found this, I can eliminate the same number x "on the columns", if it is a candidate in some other cell.

Same procedure starting to scan the rows, instead of the lines.

Hope that it helped,
see u,
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Mon Aug 29, 2005 3:30 pm    Post subject: Swordfish formation Reply with quote

OK, I see exactly how this works, now. Here's how it applies in the puzzle we started with. After 36 moves I had arrived at this position:
Code:

2/7/9     5       1      4    2/7/9    8     2/9     3      6
  6    *2/7/9*  2/7/9    1      3      5      8   *2/7/9*   4
  4       8       3    *7/9*    6    2/7/9    5   *2/7/9*   1

  3      2/9      5      6      8      4      7      1     2/9
  8       6     2/7/9    3     7/9     1      4      5     2/9
 7/9      1       4      2      5     7/9     6      8      3

2/7/9     3     2/7/9    5   1/2/7/9   6    1/2/9    4      8
  1   *2/4/7/9*   8    *7/9*  2/4/7  2/7/9    3      6      5
  5      2/4      6      8   1/2/4/9   3    1/2/9   2/9     7


(Note that there was a small error in my first post about this puzzle -- I had failed to eliminate the possibility "7" in r9c2. In this table I've also expanded all the 4-way possibilities.)

In this table, the "swordfish" consists of the six cells marked by *. I don't know yet where the 7's go in this formation, but I can be certain that either

r2c2=7 and r8c4=7 and r3c8=7

or else

r8c2=7 and r3c4=7 and r2c8=7.

Either way there are only two possible cells for the 7 in rows 2, 3, and 8. After I eliminate the other 7's marked as possible in those rows I arrive at this setup:
Code:

2/7/9     5       1      4    2/7/9    8     2/9     3      6
  6    *2/7/9*   2/9     1      3      5      8   *2/7/9*   4
  4       8       3    *7/9*    6     2/9     5   *2/7/9*   1

  3      2/9      5      6      8      4      7      1     2/9
  8       6     2/7/9    3     7/9     1      4      5     2/9
 7/9      1       4      2      5     !7!     6      8      3

2/7/9     3     2/7/9    5   1/2/7/9   6    1/2/9    4      8
  1   *2/4/7/9*   8    *7/9*   2/4    2/9     3      6      5
  5      2/4      6      8   1/2/4/9   3    1/2/9   2/9     7


The triplet {2, 7, 9} that had previously appeared in column 6 has now been resolved into a {2, 9} pair in r3c6 & r8c6, and a naked 7 that must appear in r6c6. This in its turn will force a 9 in r6c1, and the rest of the puzzle is fairly simple to work out.

Pretty slick! dcb

(BTW, I notice that the system has identified a couple of my posts as being from "guest." I don't quite understand what's up with that, because I always log on when I visit this forum. Must be a bug ...)
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someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Mon Aug 29, 2005 3:46 pm    Post subject: Reply with quote

Yup, you got it!
That is it!
That was it!
That will be!

P.S. you are sometimes a guest, because KGB can't update your logon record ;-)

see u,
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Keith
Guest





PostPosted: Sun Sep 04, 2005 8:09 pm    Post subject: Swordfish Reply with quote

I rather quickly got to the same as David's 36th move, but then I was totally stuck!

I think I see what this swordfish thing is. I'll have to think about it some more.

The Sudoku Drawer could not solve this one. Does that make it an invalid puzzle or not?

Keith
near Detroit, Michigan, USA
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samgj
Site Admin


Joined: 17 Jul 2005
Posts: 106
Location: Cambridge

PostPosted: Sun Sep 04, 2005 9:34 pm    Post subject: Re: Swordfish Reply with quote

Keith wrote:
I rather quickly got to the same as David's 36th move, but then I was totally stuck!

I think I see what this swordfish thing is. I'll have to think about it some more.

The Sudoku Drawer could not solve this one. Does that make it an invalid puzzle or not?

Keith
near Detroit, Michigan, USA


Hi Keith

The drawer currently doesn't know about swordfish things. Clearly it should, and I'll attempt to get it going as soon as I can. There's a post here about "valid" vs "invalid" puzzles. I'm persuaded that it's good to avoid calling things "invalid" just because they don't fit a personal choice of puzzle type.

Sam
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Keith
Guest





PostPosted: Mon Sep 19, 2005 2:51 am    Post subject: Re: Swordfish Reply with quote

Sam,

I think that a "valid" puzzle is one that has a single unique solution, given the starting point. Yes, there are automated Sudoku solvers that characterize puzzles as invalid or unsolvable. They seem to be making statements on their own programming, rather than on the particular puzzle at hand.

On this particular puzzle, I still do not understand the "swordfish" thing. But, I did solve the puzzle.

It comes down to R1C7: Is it a 2 or a 9?

If it is a 2, then R1C1 and R1C5 are 7 or 9. There is a cycle of cells that are only alternating values of 7 or 9: R5C5, R6C6, R6C1, and then R1C1 and R1C5. Start with any cell, with either value: 7 or 9, 9 or 7, ... You will get back to the starting cell with a contradictoryvalue.

So, R1C7 must be a 9. The next thing to test is the values in the cycle mentioned above. Pick one, fill in the rest. It is easy to see that (for example) that R5C5 is 9 (not 7), and the remainder of the puzzle is easy to solve.

Best wishes,

Keith

==================


I rather quickly got to the same as David's 36th move, but then I was totally stuck!

I think I see what this swordfish thing is. I'll have to think about it some more.

The Sudoku Drawer could not solve this one. Does that make it an
invalid puzzle or not?

Keith
near Detroit, Michigan, USA[/quote]

Hi Keith

The drawer currently doesn't know about swordfish things. Clearly it should, and I'll attempt to get it going as soon as I can. There's a post here about "valid" vs "invalid" puzzles. I'm persuaded that it's good to avoid calling things "invalid" just because they don't fit a personal choice of puzzle type.

Sam[/quote]
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